An alternative technique for fi nding the cumulative infl uence of several forces is the vector polygon method. The resultant of a force system can be found by sketching a polygon to represent the addition of the Fi vectors. The magnitude and direction of the resultant are determined by applying rules of trigonometry to the polygon’s geometry. Referring to the mounting post of Figure 4.5(a), the vector polygon for those three forces is drawn by adding the individual Fi’s in a chain according to the head-to-tail rule.
In Figure 4.5(b), the starting point is labeled on the drawing, the three forces are summed in turn, and the endpoint is labeled. The order in which the forces are drawn on the diagram does not matter insofar as the fi nal result is concerned, but diagrams will appear visually different for various addition sequences. The endpoint is located at the tip of the last vector added to the chain. As indicated in Figure 4.5(b), the resultant R extends from the start of the chain to its end. The action of R on the bracket is entirely equivalent to the combined effect of the three forces acting together. Finally, the magnitude and direction of the resultant are determined by applying trigonometric identities to the polygon’s shape. Some of the relevant equations for right and oblique triangles are reviewed in Appendix B.
We can often obtain reasonably accurate results by summing vectors on a drawing that is made to scale, for instance, 1 in. on the drawing corresponds to 100 lb. Such drafting tools as a protractor, scale, and straightedge should be used to construct the polygon and to measure the magnitudes and directions of unknown quantities. It is certainly acceptable to use a purely graphical approach when solving engineering problems, provided that the drawing is large and precise enough that you can determine the answer to a fair number of signifi cant digits.
Head-to-tail rule
Scale drawings
The eyebolt is fastened to a thick base plate, and it supports three steel cables with tensions 150 lb, 350 lb, and 800 lb. Determine the resultant force that acts on the eyebolt by using the vector algebra approach. The unit vectors i and j are oriented with the x-y coordinates as shown. (See Figure 4.6.) Example 4.1 Cable Tie-Down
4.3 Resultant of Several Forces 137
Approach
We are tasked to fi nd the resultant force on the eyebolt. By using Equations (4.1) and (4.2), we will break each force down into its horizontal and vertical components and write them in vector form. Then we will add the respective components of the three forces to fi nd the resultant’s components. Given those, the magnitude and angle of action of R follow from Equation (4.7).
Solution
The components of the 800-lb force are
Fx,1 5 (800 lb) cos 45° ← [Fx 5 F cos ] 5 565.7 lb
Fy,1 5 (800 lb) sin 45° ← [Fy 5 F sin ] 5 565.7 lb
and F1 is written in vector form as
F1 5 565.7i 1 565.7j lb ← [F 5 Fxi 1 Fyj]
By using the same procedure for the other two forces, F2 5 2(350 sin 20°)i 1 (350 cos 20°)j lb
5 2119.7i 1 328.9j lb F3 5 2150i lb
Example 4.1 continued
Figure 4.6
800 lb 350 lb
150 lb
20°
45°
i j y
x F1
F2
F3
To calculate the components of the resultant, the horizontal and vertical components of the three forces are summed separately:
Rx 5 565.7 2 119.7 2 150 lb ←
3
Rx 5S
i51N Fx,i4
5 296.0 lb
Ry 5 565.7 1 328.9 lb ←
3
Ry 5S
i51N Fy,i4
5 894.6 lb
The magnitude of the resultant force is
R 5 冑(296.0 lb)2 1 (894.6 lb)2 ←
3
R 5冑
R x2 1 R y24
5 942.3 lb and it acts at the angle
5 tan1
(
894.6 lb _______296.0 lb
)
←3
5 tan1(
___ RRxy
)
4
5 tan1
(
3.022 lb __lb
)
5 tan1(3.022)
5 71.698
which is measured counterclockwise from the x-axis. (See Figure 4.7.) Example 4.1 continued
Figure 4.7
942.3 lb 71.69°942.3 lb
296.0 lb 894.6 lb
4.3 Resultant of Several Forces 139
Discussion
The resultant force is larger than any one force, but less than the sum of the three forces because a portion of Fx,3 cancels Fx,1.The resultant force acts upward and rightward on the bolt which is expected. The three forces acting together place the bolt in tension by pulling up on it, as well as bending it through the sideways loading Rx. As a double-check on dimensional consistency when calculating the resultant’s angle, we note that the pound units in the argument of the inverse tangent function cancel. Since Rx and Ry are positive, the tip of the resultant vector lies in the fi rst quadrant of the x-y plane.
R 5 942.3 lb
5 71.69° counterclockwise from the x-axis Example 4.1 continued
The 10-lb and 25-lb forces are applied to the control lever of a mechanism.
Determine the magnitude and direction of the resultant by using the vector polygon approach. (See Figure 4.8.)
Approach
We are tasked to fi nd the resultant force on the control lever. We fi rst assume that the weight of the lever is negligible relative to the applied forces. We then sketch a vector polygon and combine the forces by using the head-to-tail rule. The two given forces, together with the resultant, will form a triangle.
We can solve for the unknown length and angle in the triangle by applying the laws of cosines and sines from Appendix B. (See Figure 4.9.)
Example 4.2 Control Lever
Figure 4.8
10 lb
25 lb 50°
Figure 4.9
Start End
40°
10 lb
25 lb R
y
x
Solution
The 25-lb force is sketched fi rst on the diagram, and the 10-lb force is added at the angle of 50° from vertical. The resultant R extends from the tail of the 25-lb force vector (which is labeled as the start point) to the head of the 10-lb force vector (the endpoint). The three vectors form a side-angle-side triangle.
Applying the law of cosines, we solve for the unknown side length R2 5 (10 lb)2 1 (25 lb)2 ← [c2 5 a2 + b2 2ab cos C]
2(10 lb)(25 lb) cos(180° 40°)
from which R 5 33.29 lb. The angle at which R acts is determined by applying the law of sines
sin(1808 2 408)______________
33.29 lb 5 sin _____
10 lb ←
3
sin A _____ a 5 sin B _____ b4
The resultant acts at 5 11.13° clockwise from the 2x-axis.
Discussion
The resultant force is less than the sum of the two forces because the forces are not aligned in the same direction. Also, the direction of the resultant force, at an angle between the two forces, matches the expectation. To double-check the solution, we could break the two forces down into their horizontal and vertical components, as in the technique in Example 4.1. By using the x-y coordinate system, the vector expression for the 10-lb force becomes F 5 27.660i 1 6.428j lb. See if you can complete the double-check for R and by using the vector algebra method.
R 5 33.29 lb
5 11.13° clockwise from the 2x-axis Example 4.2 continued
4.4 M OMENT OF A F ORCE
When you are trying to loosen a bolt, it is more easily turned if you use a wrench with a long handle. In fact, the longer the handle, the less force you need to apply with your hand. The tendency of a force to make an object rotate is called a moment. The magnitude of a moment depends both on the force applied and on the lever arm that separates the force from the point of rotation.